[Leetcode] Word Search

最新推荐文章于 2024-10-24 23:48:44 发布

原创最新推荐文章于 2024-10-24 23:48:44 发布 · 435 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 专栏收录该内容

67 篇文章

订阅专栏

本文介绍了一个算法，用于在二维字符网格中查找特定单词是否存在。通过深度优先搜索(DFS)策略，该方法遍历网格，检查每个单元格是否能构成目标单词，并确保同一单元格不被重复使用。

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,

word = "ABCB", -> returns false.

将已经search过的几位用"."标记，最后没有恢复那些标记，太懒了。。。

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        for(int i=0;i<board.size();i++)
        {
            for(int j=0;j<board[0].size();j++)
            {
                if(dfs(board,word,i,j,0,word.length()))
                {
                    return true;
                }
            }
        }
        return false;
    }
    
    bool dfs(vector<vector<char> > &board, string word, int i, int j, int depth, int maxDepth)
    {
        if(depth == maxDepth)
        {
            return true;
        }
        
        if(i<0||j<0) return false;
        if(i>=board.size()||j>=board[0].size()) return false;
        
        
        if(board[i][j]!=word[depth])
        {
            return false;
        }
        else
        {
            board[i][j] = '.';
            if(dfs(board,word,i+1,j,depth+1,maxDepth)) return true;
            if(dfs(board,word,i-1,j,depth+1,maxDepth)) return true;
            if(dfs(board,word,i,j+1,depth+1,maxDepth)) return true;
            if(dfs(board,word,i,j-1,depth+1,maxDepth)) return true;
            board[i][j] = word[depth];
            return false;
        }
    }
};